Physics+ Newton Laws, Problem 1: UNIZOR.COM - Classic Physics+ - Laws of...

Notes to a video lecture on UNIZOR.COM

Laws of Newton - Problem 1

Problem A

Prove that gravitational force of a point mass is conservative.
That is, prove that the work performed by the force of gravity of a point mass onto an object moving along a trajectory from point A to point B depends only on positions of these endpoints and independent of the trajectory an object moves along.

Proof

As was proven in an earlier lecture Field, Potential of this chapter of the course, to prove that a force is conservative, it is sufficient to show that the force is a negative gradient of some scalar function called potential.

According to the Newton's Law of Gravitation, the vector of the force of gravity produced by a point mass M and attracting a point mass m positioned on a distance r from mass M is directed along the line connecting them towards a mass M, and its magnitude equals to
F = G·M·m/r²

Let's define Cartesian coordinates with a center at a point mass M.
Coordinates of point mass M is (0,0,0).
Vector r = ||x,y,z|| represents a position of a point mass m in this system.
Now we can express the force as a vector in this system using the fact that vector r/r represents a unit vector directed from mass M to mass m.
F = −(G·M·m)·(r/r³)
In this formula we have added to the magnitude of force a multiplier r/r that represents the unit radial vector directed from M to m and the minus sign to change the direction of the vector towards mass M because the gravity attracts.

Now we will define a scalar function, the gradient of which equals to the vector of gravitational force.
Consider for now only a variable part of the vector of force r/r³.
Its representation in coordinate form is ||x/r³,y/r³,z/r³||, where r=(x²+y²+z²)^½.

Let's define a function
R(x,y,z) = R(r) =
= 1/r=1/(x²+y²+z²)^½.
Its gradient
∇R = ||∂R/∂x,∂R/∂x,∂R/∂z||
can be explicitly calculated as
∂R/∂x = −½(x²+y²+z²)^−3/2·2x =
= −x/(x²+y²+z²)³ = −x/r³
Analogously,
∂R/∂y = −½(x²+y²+z²)^−3/2·2y =
= −y/(x²+y²+z²)³ = −y/r³
∂R/∂z = −½(x²+y²+z²)^−3/2·2z =
= −z/(x²+y²+z²)³ = −z/r³

Comparing this with an expression for the force of gravity, we see that the difference between vector of force
F = −(G·M·m)·(r/r³)
and the gradient of defined above function R(x)
∇R=||−x/r³,−x/r³,−x/r³||=−r/r³
is only a constant multiplier.
Hence, F = (G·M·m)·∇R

Therefore, scalar function
U(r) = −(G·M·m)·R(r) =
= −(G·M·m)/r
where r=(x²+y²+z²)^½
has the required property, its negative gradient equals to a vector of the gravitational force.
From this follows that the force of gravity is conservative.

Note 1
This function depends not only on the field properies (mass M of the source of the field and distance r from it), but also linearly depends on a property of another object (mass m).
To make a concept of potential a property of the field only, the function U(r) is called a field potential when mass m is a unit of mass, in which case
U(r) = −G·M/r

Note 2
We can prove that gravity is a conservative force directly by following the same logic we used to prove that, if the force can be represented by a gradient of a potential, the work performed by this force is independent of the trajectory.

Consider now any two points in space A and B and some trajectory that point mass m takes to move from A to B.
The work done by any force on an object moving along some path consists of all small amounts of work the force performs on any small piece of trajectory and, by definition, equals to an infinite sum (that is, integral along a path) of infinitesimal increments *that is, differentials) of work, each of which is a scalar product of the vector of force F and infinitesimal vector of the increment of position along a trajectory dr
W_[AB] = ∫_[AB]dW = ∫_[AB]F·dr

To calculate the scalar product F·dr, we can express both in coordinate form
F = −(G·M·m/r³)·||x,y,z|| =
= −(G·M·m/r³)·r
dr = ||dx,dy,dz||
F·dr = −(G·M·m/r³)·
·(x·dx+y·dy+z·dz) - a scalar

Notice that
x·dx + y·dy + z·dz =
d(x²/2 + y²/2 + z²/2) =
= (1/2)d(r²) = r·dr

Therefore,
F·dr = −(G·M·m/r³)·r·dr =
= −(G·M·m/r²)·dr =
= −(G·M·m)·d(1/r) =
= d(−G·M·m/r)

Since F·dr is a full differential of some function, integral of it along a path from A to B equals to a difference of the values of this function at the points of limits of integration
∫_[AB]F·dr = ∫_[AB]d(−G·M·m/r) =
= −G·M·m/r(B)+G·M·m/r(A) =
= G·M·m/r(A) − G·M·m/r(B)
which proves that work performed by the force of gravity on an object moving along some trajectory is independent of the trajectory, that is it proves that gravitational force is conservative.

Physics+ Potential Theorem: UNIZOR.COM - Classic Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Potential Theorem

As you recall, we have defined a field as an area (a subset) of points in our three-dimensional space with a vector of force F(x,y,z) called field intensity defined at each point of this area and a real scalar function U(x,y,z) called potential defined at all these points, when the following equation between a force and a potential at each point P(x,y,z) is held:
F(x,y,z) = −∇U(x,y,z)
where symbol ∇ signifies gradient of a function U(x,y,z) - a vector of function's partial derivatives by each coordinate
∇U(x,y,z) =
= ||∂U/∂x, ∂U/∂y, ∂U/∂z||

We have also proven that the work performed by such a field intensity function on an object moving along some trajectory depends only on the endpoints of the object's movement and is independent of a path chosen between these two endpoints.
Thus, independence of work of the trajectory is a necessary condition of the existence of the field potential function U(x,y,z), whose gradient with a minus sign equals to the field intensity force F(x,y,z).

In this lecture we will prove that the condition of work being independent of a trajectory between the endpoints of an object's movement is also a sufficient condition for the field force to be a gradient of some scalar function - the field potential.

Theorem

A vector of force F(x,y,z) is defined at each point P(x,y,z) of a certain area within our three-dimensional Euclidean space with Cartesian coordinates.

An object, acted upon by this force, moves within this area.

It's given that the work performed by this force on an object moving along some trajectory between any two points depends only on positions of these endpoints and is independent of the trajectory between them.

Prove that there exists a scalar function of position U(x,y,z) called potential such that the vector of force equals to a negative gradient of this potential, that is
F(x,y,z) = −∇U(x,y,z) =
= −||∂U/∂x, ∂U/∂y, ∂U/∂z||

Since vector F(x,y,z) at point P(x,y,z) can be expressed in coordinate form as
||F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)||
the above statement can be formulated in coordinate form as
F_x(x,y,z) = −∂U(x,y,z)/∂x
F_y(x,y,z) = −∂U(x,y,z)/∂y
F_z(x,y,z) = −∂U(x,y,z)/∂z

Proof

As a proof, we will explicitly define a potential function and prove that it satisfies the required equalities.

Choose arbitrarily some fixed point A(x₀,y₀,z₀) in the area where our force is defined, that we will use as the beginning of some trajectory.
Choose any other point B(x,y,z) there, where we will explicitly define the scalar function U(x,y,z) (the potential) that satisfies the conditions of the problem.

Let's chose any particular path from point A(x₀,y₀,z₀) to point B(x,y,z) and define the function U(x,y,z) (the field potential) for point B(x,y,z) as the work of force F performed during an object's movement along a chosen path from A(x₀,y₀,z₀) to B(x,y,z) with a negative sign.
This definition of a potential is quite legitimate since the work performed by force F does not depend on a path chosen, but depends only on position of points A and B.
So, by definition, for any point B(x,y,z) where the field is defined
U(B) = U(x,y,z) = −W_[AB]

Let's chose a point C(x+dx,y+dy,z+dz) on infinitesimal distance from point B(x,y,z).
Let dr be a vector of displacement from B to C:
dr = ||dx,dy,dz||

According to the definition of work, when the field intensity force F acts on an object that moves from point B(x,y,z) to an infinitesimally close to it point C(x+dx,y+dy,z+dz), the infinitesimal amount of work performed by the force is equal to
dW_[BC] = F(x,y,z)·dr =
= F_x(x,y,z)·dx + F_y(x,y,z)·dy +
+ F_z(x,y,z)·dz

At the same time, using the Lemma D of the previous lecture Work Lemmas, this same amount of work equals to
dW_[BC] = W_[AC] − W_[AB]
where fixed point A(x₀,y₀,z₀) was arbitrarily chosen above.

Expressions W_[AB] and W_[AC] represent the amount of work the field force performs, if an object moves from point A to B and from points A to C correspondingly. These same amounts were used to define a potential at these point
W_[AB] = −U(x,y,z)
W_[AC] = −U(x+dx,y+dy,z+dz).

Therefore,
dW_[BC]= −U(x+dx,y+dy,z+dz)+
+ U(x,y,z) = −dU(x,y,z)
where dU(x,y,z) is a full differential (infinitesimal increment) of function U(x,y,z) on an interval from B(x,y,z) to C(x+dx,y+dy,z+dz).

As known from Calculus, the full differential of a function can be expressed in terms of partial derivatives and differentials of arguments (see Partial Derivatives - Basic Properties lecture of the Calculus chapter in the course Math 4 Teens on UNIZOR.COM).
Therefore,
dW_[BC] = −dU(x,y,z) =
= −(∂U(x,y,z)/∂x)·dx −
− (∂U(x,y,z)/∂y)·dy −
− (∂U(x,y,z)/∂z)·dz

Comparing this expression of dW_[BC] with the one in terms of the field force components above, we come to an equation
F_x(x,y,z)·dx + F_y(x,y,z)·dy +
+ F_z(x,y,z)·dz =
= −(∂U(x,y,z)/∂x)·dx −
− (∂U(x,y,z)/∂y)·dy −
− (∂U(x,y,z)/∂z)·dz

While dx, dy and dz are infinitesimal increments of position along some infinitesimal displacement for each coordinate, the direction of this displacement can be chosen freely.
Choosing infinitesimal dx and dy=0, dz=0 leads to an equality
F_x(x,y,z) = −∂U(x,y,z)/∂x
Similarly, leaving only dy or dz as infinitesimal increments and setting displacement along other coordinates to 0, we obtain the equalities
F_y(x,y,z) = −∂U(x,y,z)/∂y
F_z(x,y,z) = −∂U(x,y,z)/∂z
End of Proof

The question now arises, is the field potential U(x,y,z) uniquely defined by the field force intensity?
The answer is NO, since we have chosen point A(x₀,y₀,z₀) as, basically, any fixed reference point where the movement of an object begins.
We can choose any other point A'(x₁,y₁,z₁) as the beginning, and the work W(x,y,z) will be different. More precisely, it will differ by the amount of work the force performs moving an object from A to A'.

This means that our field potential is not uniquely defined by the field force, only its partial derivatives are, since they must correspond to force components. This is similar to the fact that, given a derivative of a function, the function is defined as an integral from a derivative plus some freely chosen constant.

Traditionally, for gravitational or electrostatic fields, as the staring point, physicists choose a point infinitely far from the source of the field force. There the force is equal to zero. With this convention the field potential U(x,y,z) is fully defined and equals to the negative work performed by the force F(x,y,z) to move an object from an infinitely far point to any point P(x,y,z).

Physics+ Work Lemmas: UNIZOR.COM - Classic Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Field Work Lemmas

In the previous lecture we have introduced the concepts of a field and field intensity force that is equal to a gradient of the field potential.
Also, we have proven that, dealing with such force, the work of this field intensity force along any trajectory of an object moving in the field depends only on the field potential at the beginning and at the end of a trajectory and is independent of a path between these two points.

There is a converse theorem that states that, if the work performed by some force on an object depends on the object's position in the beginning and at the end of its movement and does not depend on a trajectory between these points, then this force can be represented as a gradient of some scalar function, the field potential.

This lecture presents certain auxiliary theorems (lemmas) that will help to prove the above mentioned theorem in the next lecture.


Lemma A

This lemma, in short, is about comparing the work performed by a force, when an object moves along the same trajectory in two opposite directions.

More rigorously, assume, some vector of force F(x,y,z) is defined at each point of a certain area within our three-dimensional Euclidean space with Cartesian coordinates.
The coordinate components of the vector of force F at point (x,y,z) are
||F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)||.

An object is moving along certain trajectory from point A to point B within this area, while the force F acts on it and performs certain work.
The coordinate components of the position vector r are
||x,y,z||.
The coordinate components of the infinitesimal increment, the differential, of the position vector dr are
||dx,dy,dz||.

Compare the work done by the force F along the object's trajectory from point A to point B with the work done when an object moves from point B to point A along the same trajectory in the opposite direction.

Solution A

Recall the definition of work performed by force F on an object moving along a trajectory described by position vector r from starting point A to finishing point B:
W_[AB] = ∫_[AB]dW = ∫_[AB]F·dr
The above integral is an infinite sum of infinitesimal work increments dW(x,y,z) performed by force F(x,y,z) on an infinitesimal interval of a trajectory r(x,y,z) from point (x,y,z) to point (x+dx,y+dy,z+dz), where F·dr is a scalar product of two vectors, so
dW(x,y,z) = F(x,y,z)·dr(x,y,z) =
= F_x·dx+F_y·dy+F_z·dz

Assume, at some moment of time our object, moving from A to B, is at position (x,y,z).
During an infinitesimal increment of time it's new position will be (x+dx,y+dy,z+dz) and the force will do an infinitesimal amount of work
dW = F_x·dx+F_y·dy+F_z·dz.

Now assume that the object moves in the opposite direction from B to A along the same trajectory.
Being at the same position (x,y,z) at some moment in time, the infinitesimal increments dx, dy and dz will have signs opposite to those when an object moved from A to B, while the force vector will be the same.

Therefore, the differential of work dW will also be of an opposite sign, and subsequent integration will result in total work done by the force on trajectory from B to A to have the same magnitude but opposite sign comparing with object movement from A to B.

Answer A
W_[AB]=∫_[AB]F·dr =
= −∫_[BA]F·dr = −W_[BA]


Lemma B

As in Lemma A, assume, some vector of force F(x,y,z) is defined at each point of a certain area within our three-dimensional Euclidean space with Cartesian coordinates.

An object is moving along certain trajectory within this area and the force F(x,y,z) acts on it, performing some work along its trajectory.
The object's position at time t is r(t)=||x(t),y(t),z(t)||.

It's given that the work of this force on an object moving along any trajectory between any pair of points A and B depends only on a choice of these two endpoints and does not depend on a choice of trajectory between them.

Prove that the work this force performs on an object moving along a closed trajectory, when points A and B coincide, equals to zero.

Proof B

In our case of a closed trajectory the endpoints A and B coincide. So, we will deal only with point A.

Choose any closed trajectory, its starting and ending point A and point M on it that does not coincide with point A.
Now we have two different paths from A to M, let's call them path #1 and path #2.

If an object moves along a closed trajectory, it moves from point A to point M along path #1 and then moves from point M to point A along path #2.

According to the condition of the problem, a work W_1[AM] performed by a force acting on our object along a path #1 from A to M should be equal to a work W_2[AM] performed by a force acting on our object along a path #2 from A to M:
W_1[AM] = W_2[AM]

As has been proven in the Lemma 1, during the second part of the trajectory, when object moves from M to A along path #2, the work of a force is of the same magnitude as if an object moved from A to M along the same path #2 but with an opposite sign:
W_2[MA] = −W_2[AM]

Therefore, the total work along path #1 from A to M followed by moving from M to A along path #2 equals to
W_1[AM]2[MA] = W_1[AM]+W_2[MA] =
= W_1[AM] − W_2[AM] =
= W_1[AM] − W_1[AM] = 0


Lemma C

This lemma is a converse to Lemma B.
As in Lemma A, assume, some vector of force F(x,y,z) is defined at each point of a certain area within our three-dimensional Euclidean space with Cartesian coordinates.

An object is moving along certain trajectory within this area, and the force F(x,y,z) acts on it, performing some work.
The object's position at time t is r(t)=||x(t),y(t),z(t)||.

It's given that the work of this force on an object moving along any closed trajectory that starts and ends at the same point equals to zero.

Prove that the work this force performs on an object moving from any fixed point A to any fixed point B does not depend on trajectory between these points.

Proof C

Choose any two paths from point A to point B - path #1 and path #2.
We will use the symbols introduced in Lemma 2.
As stated in the condition of this lemma,
W_1[AB]2[BA] =
= W_1[AB] + W_2[BA] = 0
At the same time
W_2[BA] = −W_2[AB]
Therefore,
0 = W_1[AB] + W_2[BA] =
= W_1[AB] − W_2[AB]
from which follows that
W_1[AB] = W_2[AB]

As we see, a statement that the work performed by a force on an object moving along any closed trajectory equals to zero is equivalent to a statement that the work performed by a force on an object moving from any point A to any point B does not depend on a trajectory an object moves between these points.


Lemma D

As in Lemma A, assume, some vector of force F(x,y,z) is defined at each point of a certain area within our three-dimensional Euclidean space with Cartesian coordinates.
It's given that the work of this force on an object moving along any trajectory between any pair of points depends only on a choice of these two endpoints and does not depend on a choice of trajectory between them.

Assume, there are three points in the area where the force is acting, A, B and C.
Consider amounts of work the force performs on an object during its movements between these points
W_[AB], W_[AC], W_[BC].

Prove that
W_[BC] = W_[AC] − W_[AB].

Proof D

From Lemma B above follows that the amount of work performed by a force on an object moving along a closed trajectory from A to B to C and back to A equals to zero
W_[AB] + W_[BC] + W_[CA] = 0

From Lemma A above follows that, if object moves in the opposite direction along the same trajectory, the amount of work performed by a force is the same in magnitude but opposite in sign to the amount of work performed on a directly moving object
W_[AC] = −W_[CA]

Therefore,
to B to C and back to A equals to zero
W_[AB] + W_[BC] − W_[AC] = 0
from which follows that
W_[BC] = W_[AC] − W_[AB].

Coincidentally, this equality reminds the rule about a difference between two vectors
BC = AC − AB

Physics+ Field, Potential: UNIZOR.COM - Classic Physics+ - Laws of Newto...

Notes to a video lecture on UNIZOR.COM

Field, Potential

The prototype for an abstract concept of a field presented below is a gravitational field. The force of this field acting on a unit mass is a prototype for a concept of field intensity.

On UNIZOR.COM these concepts were introduced in the Physics 4 Teens course, part Energy, chapters Energy of Gravitational Field and Gravitational Potential.
Here we will formally define these and other concepts and derive a few important field properties.

Field is an area (a subset) of points P{x,y,z} in our three-dimensional space with a vector of force F(P) called field intensity defined at each point P of this area and a real function U(P) called potential defined at exactly the same points, when the following equation between a force and a potential at each point is held:
∀P{x,y,z}: F(P) = −∇U(P)
where symbol ∇ signifies gradient of a function U - a vector of function's partial derivatives by each coordinate ∇U(P) = ∇U(x,y,z) =
= ||∂U/∂x,∂U/∂y,∂U/∂z||

It should be noted that in many cases authors do not differentiate between the field and the field intensity force, defining the field as the force, which in our opinion is misleading.
That's why we define field as an area of a space, which corresponds to the usual meaning of this word, and field intensity as a force acting inside this area.

Let's assume that a material point acted upon by the force of a field is moving during the time t period from t=t₁ to t=t₂ from point A to point B along some trajectory
P(t) = {x(t),y(t),z(t)}
where P(t₁)=A and P(t₂)=B

The first important property of a field is that the work of the field intensity force F(P) along any trajectory of an object moving in the field depends only on the field potential at the beginning and at the end of a trajectory and is independent of a path between these two points. In other words, no matter how an object that experiences a field force moves from point A to point B, the work of the force remains the same and depends only on the field potential at end points A and B.

Here is the proof of this statement.
Assume, at moment t₁ our object is at point A, and it moves along a trajectory r(t)=||x(t),y(t),z(t)|| until at moment t₂ it reaches point B.
The field intensity F is defined for all points of a field, including the points along the trajectory of an object and is equal to F(t)=F(x(t),y(t),z(t)).

Work performed by any force during the observed time is, by definition,
W_t∈[t1,t2] = ∫_t∈[t1,t2]F(t)·dr(t)
In our case of a field force
F(t) = F(x(t),y(t),z(t)) =
= −∇U(x(t),y(t),z(t)) =
= −||∂U/∂x,∂U/∂y,∂U/∂z||
and
dr(t)=||dx(t),dy(t),dz(t)||

Let's evaluate the scalar product of two vectors under the integral
||∂U/∂x,∂U/∂y,∂U/∂z||·
||dx(t),dy(t),dz(t)|| =
= (∂U/∂x)·dx(t) +
+ (∂U/∂y)·dy(t) +
+ (∂U/∂z)·dz(t) =
= dU(x(t),y(t),z(t)) = dU(t)
Above is a full differential (infinitesimal increment) of function U(t)=U(x(t),y(t),z(t)) on infinitesimal time interval from t to t+dt.

Therefore, the work performed by a field force F(t) equals to
W_t∈[t1,t2] = −∫_t∈[t1,t2]dU(t) =
= −U(t₂) + U(t₁) = U(t₁) − U(t₂)
As we see, the total amount of work performed by a field intensity force depends only on the field potentials at end points and does not depend on the path (trajectory) an object took to reach from start to finish.

An obvious consequence of this property of the field is that an amount of work the field intensity force performed when an object moves along a trajectory with the finishing point coinciding with the starting one equals to zero.

Recall that amount of work the field intensity force performed when an object moves from one point to another equals to an increment of the object's kinetic energy (see the previous lecture Newton's Laws of this chapter of a course)
W_t∈[t1,t2] = T(t₂) − T(t₁)
Therefore,
T(t₂) − T(t₁) = U(t₁) − U(t₂)
from which follows the Law of Conservation of Energy
T(t₁) + U(t₁) = T(t₂) + U(t₂)
It states that the sum of kinetic and potential energy is not changing during an object's movement within a field.